a)(1/2 + 1/6 + 1/12 +...+1/90) . 100 - [5/2 : ( x + 206/100 )] : 1/2= 89
b) x9,85 + x0,15 = 0,1
c) 2/5 + 2022x = 4/10
giúp mình, 2h mình đi học r
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\(a)\) \(2x-5=21\)
\(\Leftrightarrow\) \(2x=21+5\)
\(\Leftrightarrow\) \(2x=26\)
\(\Leftrightarrow\) \(x=26:2\)
\(\Leftrightarrow\) \(=13\)
\(b)\) \(\frac{3}{4}+\frac{1}{4}x=\frac{5}{6}\)
\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{5}{6}-\frac{3}{4}\)
\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{1}{12}\)
\(\Leftrightarrow\) \(x=\frac{1}{3}\)
c)
\(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+....+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\)
\(\left(1+1+1+....+1+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{6\times7}+\frac{1}{7\times8}\right)\)(Có 7 số 1)
\(7+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(7+1-\frac{1}{8}=\frac{63}{8}\)
Gợi ý 1 bài c) còn d) e) cũng làm như vậy nhé
Chúc bạn học tốt !!!
a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-...+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\dfrac{9}{10}\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=90-89\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=1\)
\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}\)
\(\Rightarrow x=\dfrac{11}{25}\)
b) \(x\cdot9,85+x\cdot0,15=0,1\)
\(\Rightarrow x\cdot\left(9,85+0,15\right)=0,1\)
\(\Rightarrow x\cdot10=0,1\)
\(\Rightarrow x=\dfrac{0,1}{10}\)
\(\Rightarrow x=0,01\)
c) \(\dfrac{2}{5}+2022x=\dfrac{4}{10}\)
\(\Rightarrow\dfrac{2}{5}+2022x=\dfrac{2}{5}\)
\(\Rightarrow2022x=\dfrac{2}{5}-\dfrac{2}{5}\)
\(\Rightarrow2022x=0\)
\(\Rightarrow x=\dfrac{0}{2022}\)
\(\Rightarrow x=0\)
a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\left(1\right)\)
Ta có :
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(\left(1\right)\Rightarrow\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=89\)
\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=90-89\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)
\(\Rightarrow x+\dfrac{206}{100}=\dfrac{5}{2}:\dfrac{1}{2}\)
\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}.\dfrac{2}{1}\)
\(\Rightarrow x+\dfrac{103}{50}=5\)
\(\Rightarrow x=5-\dfrac{103}{50}\)
\(\Rightarrow x=\dfrac{250}{50}-\dfrac{103}{50}\)
\(\Rightarrow x=\dfrac{147}{50}\)